{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Practice Test 3 solutions

# Practice Test 3 solutions - ISyE 3232C Stochastic...

This preview shows pages 1–2. Sign up to view the full content.

ISyE 3232C Stochastic Manufacturing and Service Systems Spring 2010 Practice Test 3 (Solution) Closed book, closed notes, no electronics. 1. Let X p be the service time for a perm, and X m be the service time for a manicure. Then, X p exp(2) and X m exp(3). Also, let T S , T L , T G be the service time of Sue, Liz and Grace, respectively. Note that Y L = X p , Y G = X m and Y S = X p . (a) E [min { Y L , Y G } ] + E [ Y S ] = 1 2 + 3 + 1 2 = 7 10 (b) P ( Y L < Y G ) P ( Y S > Y G ) + P ( Y L > Y G ) P ( Y S > Y L ) = 2 2 + 3 3 2 + 3 + 3 2 + 3 2 2 + 2 = 27 50 (c) P ( Y L > Y G ) P ( Y L > Y S ) = 3 2 + 3 2 2 + 2 = 3 10 2. Let X A be the remaining lifetime of person A , and Y be the interarrival time of kidney arrivals. Then, X A exp( μ A ) and Y exp( λ ). Then, the probability that A obtains a new kidney is same as P ( X A > Y ) = λ λ + μ A . 3. { N ( t ) : t 0 } is a P.P( λ = 30/hour). (a) E [ S 4 ] = 4 30 (b) E [ N (7) - N (5) | N (4) - N (3) = 3] = E [ N (7) - N (5)] = E [ N (2) - N (0)] = E [ N (2)] = 2 × 30 = 60. (c) { N 1 ( t ) } is a P.P( λ 1 = 24/hour) of car arrivals, and { N 2 ( t ) } is a P.P( λ 2 = 6/hour) of van arrivals.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

Practice Test 3 solutions - ISyE 3232C Stochastic...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online